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Hospital's Rule: Statement and Proof

What is Hospital's theorem, also known as Hospital's rule? This article will make it possible to state and demonstrate this rule!

Hospital rule statement

Let f and g be two differentiable functions at a point a and such that f(a) = g(a) = 0. Moreover, g'(a) ≠ 0. Then, we have the following property:

\lim_{x \to a} \dfrac{f(x)}{g(x)} = \lim_{x \to a } \dfrac{f'(x)}{g'(x)}


We can replace a by

\ pm \ infty

And also we can have:

\begin{array}{l} \displaystyle\lim_{x\to a} f(x) = \pm \infty\\ \displaystyle\lim_{x\to a} g(x) = \pm \infty \end {array}


Example 1 : Take the function h defined by

h(x) = \dfrac{x-2}{x^2-6x+8} = \dfrac{f(x)}{g(x)} 

We have :


On the other hand,

g'(x) = 2x-6


g'(2) \neq 0 

We then deduce:

\lim_{x \to 2} h(x) = \lim_{x \to 2} \dfrac{1}{2x-6} = -\dfrac{1}{2}

Example 2 : Take the functions f and g defined by

f(x) = \ln(X)


g(x) = x-1

And let's study the limit of the ratio of these two functions in 1.
We have: f(1) = g(1) = 0
Urban artist

g'(x) = 1 

So we also have:

g'(1) \neq 0 

We will therefore apply our famous rule to obtain:

\lim_{x \to 1}\dfrac{f(1)}{g(1)} = \lim_{x \to 1}\dfrac{f'(1)}{g'(1)} = \dfrac {\frac{1}{1}}{1} = 1

Demonstration of L'Hôpital's rule

Now let's move on to the demonstration. This is quite simple in the classical case.
We have the following property:

f(x) = f(x)-f(a)


g(x) =g(x)-g(a)

So we do the quotient:

\begin{array}{ll} \dfrac{f(x)}{g(x)} & = \dfrac{f(x)-f(a)}{g(x)-g(a) }\\ & = \dfrac{f(x)-f(a)}{xa } \dfrac{xa}{g(x)-g(a) }\\ \end{array}

Passing to the limit:

\lim_{x\to a }\dfrac{f(x)}{g(x)} =\dfrac{f'(a)}{g'(a)}

This concludes this demonstration of L'Hôpital's rule.

If you are familiar with the usual equivalents and expansions, you will normally never need this rule. But sometimes it can be easy to use and some are fans of it as they master the limited developments!


  1. Good evening sir. Is this an expansion limited to order 1 that you used to prove L'Hôpital's rule? If so I think that instead of f'(x) it's f'(a) (Taylor Young formula). Please enlighten me on this. Thanks in advance

    • Hi,
      No expansion limited to order 1: Once we know the expansions limited, the rule of the hospital becomes useless. Limited developments are easier to use.
      Following your remarks, I decided to use another demonstration, which seems simpler to me.
      Good evening and available for your questions,

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